Simplify and expand the following expression: $ \dfrac{4}{5q - 5}+ \dfrac{3}{4q - 36}- \dfrac{3q}{q^2 - 10q + 9} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{4}{5q - 5} = \dfrac{4}{5(q - 1)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{3}{4q - 36} = \dfrac{3}{4(q - 9)}$ We can factor the quadratic in the third term: $ \dfrac{3q}{q^2 - 10q + 9} = \dfrac{3q}{(q - 1)(q - 9)}$ Now we have: $ \dfrac{4}{5(q - 1)}+ \dfrac{3}{4(q - 9)}- \dfrac{3q}{(q - 1)(q - 9)} $ The least common multiple of the denominators is: $ 20(q - 1)(q - 9)$ In order to get the first term over $20(q - 1)(q - 9)$ , multiply by $\dfrac{4(q - 9)}{4(q - 9)}$ $ \dfrac{4}{5(q - 1)} \times \dfrac{4(q - 9)}{4(q - 9)} = \dfrac{16(q - 9)}{20(q - 1)(q - 9)} $ In order to get the second term over $20(q - 1)(q - 9)$ , multiply by $\dfrac{5(q - 1)}{5(q - 1)}$ $ \dfrac{3}{4(q - 9)} \times \dfrac{5(q - 1)}{5(q - 1)} = \dfrac{15(q - 1)}{20(q - 1)(q - 9)} $ In order to get the third term over $20(q - 1)(q - 9)$ , multiply by $\dfrac{20}{20}$ $ \dfrac{3q}{(q - 1)(q - 9)} \times \dfrac{20}{20} = \dfrac{60q}{20(q - 1)(q - 9)} $ Now we have: $ \dfrac{16(q - 9)}{20(q - 1)(q - 9)} + \dfrac{15(q - 1)}{20(q - 1)(q - 9)} - \dfrac{60q}{20(q - 1)(q - 9)} $ $ = \dfrac{ 16(q - 9) + 15(q - 1) - 60q} {20(q - 1)(q - 9)} $ Expand: $ = \dfrac{16q - 144 + 15q - 15 - 60q}{20q^2 - 200q + 180} $ $ = \dfrac{-29q - 159}{20q^2 - 200q + 180}$